Question: A particle moves along the curve $xy^3=40$ so that the $x$ -coordinate is increasing at a constant rate of $5$ units per minute. What is the rate of change (in units per minute) of the particle's $y$ -coordinate when the particle is at the point $(5,2)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-6$ (Choice B) B $4$ (Choice C) C $-\dfrac{2}{3}$ (Choice D) D $\dfrac13$
Answer: Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This, however, shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dx}{dt}=5$ for any value of $t$. We are asked for the rate of change of the particle's $y$ -coordinate when the particle is at the point $(5,2)$. In other words, we are asked for the value of $\dfrac{dy}{dt}$ at the point $(5,2)$. Finding $\dfrac{dy}{dt}$ $\dfrac{dy}{dt}=-\dfrac{5y}{3x}$ Finding $\dfrac{dy}{dt}$ at $(5,2)$ The expression for $\dfrac{dy}{dt}$ depends on both the particle's $x$ -coordinate ${5}$ and its $y$ -coordinate ${2}$ : $\begin{aligned} \dfrac{dy}{dt}&=-\dfrac{5({2})}{3({5})} \\\\ &=-\dfrac23 \end{aligned}$ In conclusion, the rate of change of the particle's $y$ -coordinate when the particle is at the point $(5,2)$ is $-\dfrac23$ units per minute.